D.80 Hy­dro­gen ground state Stark ef­fect

This note de­rives the Stark ef­fect on the hy­dro­gen ground state. Since spin is ir­rel­e­vant for the Stark ef­fect, it will be ig­nored.

The un­per­turbed ground state of hy­dro­gen was de­rived in chap­ter 4.3. Fol­low­ing the con­ven­tion in per­tur­ba­tion the­ory to ap­pend a sub­script zero to the un­per­turbed state, it can be sum­ma­rized as:

\begin{displaymath}
H_0\psi_{100,0} = E_{100,0}\psi_{100,0}
\qquad H_0 = -\fra...
...^2+V
\qquad \psi_{100,0}=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}
\end{displaymath}

where $H_0$ is the un­per­turbed hy­dro­gen atom Hamil­ton­ian, $\psi_{100,0}$ the un­per­turbed ground state wave func­tion, $E_{100,0}$ the un­per­turbed ground state en­ergy, 13.6 eV, and $a_0$ is the Bohr ra­dius, 0.53 Å.

The Stark per­tur­ba­tion pro­duces a change $\psi_{100,1}$ in this wave func­tion that sat­is­fies, from (A.243),

\begin{displaymath}
(H_0-E_{100,0})\psi_{100,1} = -(H_1-E_{100,1})\psi_{100,0}
\quad H_1 = e{\cal E}_{\rm {ext}}z
\end{displaymath}

The first or­der en­ergy change $E_{100,1}$ is zero and can be dropped. The so­lu­tion for $\psi_{100,1}$ will now sim­ply be guessed to be $\psi_{100,0}$ times some spa­tial func­tion $f$ still to be found:

\begin{displaymath}
\left(H_0-E_{100,0}\right)(f\psi_{100,0}) =
-e{\cal E}_{\r...
...si_{100,0}
\qquad H_0 = -\frac{\hbar^2}{2m_{\rm e}}\nabla^2+V
\end{displaymath}

Dif­fer­en­ti­at­ing out the Lapla­cian $\nabla^2$ of the prod­uct $f\psi_{100,0}$ into in­di­vid­ual terms us­ing Carte­sian co­or­di­nates, the equa­tion be­comes

\begin{displaymath}
f \left(H_0-E_{100,0}\right)\psi_{100,0}
- \frac{\hbar^2}{...
...(\nabla^2 f)\psi_{100,0}
=
-e{\cal E}_{\rm ext}z\psi_{100,0}
\end{displaymath}

The first term in this equa­tion is zero since $H_0\psi_{100,0}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{100,0}\psi_{100,0}$. Also, now us­ing spher­i­cal co­or­di­nates, the gra­di­ents are, e.g. [40, 20.74, 20.82],

\begin{displaymath}
\nabla f = \frac{\partial f}{\partial r} {\hat\imath}_r +
...
...bla \psi_{100,0} = - \psi_{100,0} \frac{1}{a_0} {\hat\imath}_r
\end{displaymath}

Sub­sti­tut­ing that into the equa­tion, it re­duces to

\begin{displaymath}
\frac{\hbar^2}{m_{\rm e}}
\left(\frac{1}{a_0}\frac{\partia...
...2 f\right)
\psi_{100,0}
=
-e{\cal E}_{\rm ext}z\psi_{100,0}
\end{displaymath}

Now $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r\cos\theta$ in po­lar co­or­di­nates, and for the $r$-​de­riv­a­tive of $f$ to pro­duce some­thing that is pro­por­tional to $r$, $f$ must be pro­por­tional to $r^2$. (The Lapla­cian in the sec­ond term al­ways pro­duces lower pow­ers of $r$ than the $r$-​de­riv­a­tive and can for now be ig­nored.) So, to bal­ance the right hand side, $f$ should con­tain a high­est power of $r$ equal to:

\begin{displaymath}
f = - \frac{m_{\rm e}e{\cal E}_{\rm ext}a_0}{2\hbar^2} r^2 \cos\theta + \ldots
\end{displaymath}

but then, us­ing [40, 20.83], the $\nabla^2f$ term in the left hand side pro­duces an $e{\cal E}_{\rm {ext}}a_0\cos\theta$ term. So add an­other term to $f$ for its $r$-​de­riv­a­tive to elim­i­nate it:

\begin{displaymath}
f = - \frac{m_{\rm e}e{\cal E}_{\rm ext}a_0}{2\hbar^2} r^2 ...
...\frac{m_{\rm e}e{\cal E}_{\rm ext}a_0^2}{\hbar^2} r \cos\theta
\end{displaymath}

The Lapla­cian of $r\cos\theta$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z$ is zero so no fur­ther terms need to be added. The change $f\psi_{100,0}$ in wave func­tion is there­fore

\begin{displaymath}
\psi_{100,1}= - \frac{m_{\rm e}e{\cal E}_{\rm ext}a_0}{2\hb...
...{\pi a_0^3}}
\left(r^2 + 2 a_0 r\right) e^{-r/a_0}\cos \theta
\end{displaymath}

(This small per­tur­ba­tion be­comes larger than the un­per­turbed wave func­tion far from the atom be­cause of the grow­ing value of $r^2$. It is im­plic­itly as­sumed that the elec­tric field ter­mi­nates be­fore a real prob­lem arises. This is re­lated to the pos­si­bil­ity of the elec­tron tun­nel­ing out of the atom if the po­ten­tial far from the atom is less than its en­ergy in the atom: if the elec­tron can tun­nel out, there is strictly speak­ing no bound state.)

Now ac­cord­ing to (A.243), the sec­ond or­der en­ergy change can be found as

\begin{displaymath}
E_{100,2} = \langle\psi_{100,0}\vert H_1\psi_{100,1}\rangle
\qquad H_1 = e{\cal E}_{\rm {ext}} r \cos \theta
\end{displaymath}

Do­ing the in­ner prod­uct in­te­gra­tion in spher­i­cal co­or­di­nates pro­duces

\begin{displaymath}
E_{100,2} = - \frac{9m_{\rm e}e^2{\cal E}_{\rm ext}^2a_0^4}{4\hbar^2}
\end{displaymath}